HDU-4717 The Moving Points(凸函数求极值)

The Moving Points

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2122    Accepted Submission(s): 884

Problem Description

There are N points in total. Every point moves in certain direction and certain speed. We want to know at what time that the largest distance between any two points would be minimum. And also, we require you to calculate that minimum distance. We guarantee that no two points will move in exactly same speed and direction.

 

Input

The rst line has a number T (T <= 10) , indicating the number of test cases.
For each test case, first line has a single number N (N <= 300), which is the number of points.
For next N lines, each come with four integers X_i, Y_i, VX_i and VY_i (-10⁶ <= X_i, Y_i <= 10⁶, -10² <= VX_i , VY_i <= 10²), (X_i, Y_i) is the position of the i^th point, and (VX_i , VY_i) is its speed with direction. That is to say, after 1 second, this point will move to (X_i + VX_i , Y_i + VY_i).

 

Output

For test case X, output "Case #X: " first, then output two numbers, rounded to 0.01, as the answer of time and distance.

 

Sample Input

2

2

0 0 1 0

2 0 -1 0

2

0 0 1 0

2 1 -1 0

 

Sample Output

Case #1: 1.00 0.00

Case #2: 1.00 1.00

 

 给出几个点的坐标和xy方向上的坐标分速度，问什么时候两点之间距离最大值最小，可以想到两点之间距离要么一直增大，要么先减小后增大，三分就可以啦

#pragma GCC diagnostic error "-std=c++11"
//#include <bits/stdc++.h>
#define _ ios_base::sync_with_stdio(0);cin.tie(0);
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

const double eps = 1e-;
const int N=  + ;
struct point{
    double x, y, vx, vy;
    void read(){ cin >> x >> y >> vx >> vy; }
}p[N];
int n;

double dist_points(point p1, point p2, double t){
    double x = (p1.x + t * p1.vx) - (p2.x + t * p2.vx);
    double y = (p1.y + t * p1.vy) - (p2.y + t * p2.vy);
    return sqrt(x * x + y * y);
}

double cal(double x){
    double Max = ;
    for(int i = ; i < n; i++)
        for(int j = i + ; j < n; j++)
            Max = max(Max, dist_points(p[i], p[j], x));
    return Max;
}

double ternary_search(double L, double R){
    if(L > R) swap(L, R);
    while(R - L > eps){
        double mid1, mid2;
        mid1 = (L + R) / ;
        mid2 = (mid1 + R) / ;
        if(cal(mid1) <= cal(mid2)) R = mid2;
        else L = mid1;
    }
    return (L + R) / ;
}
int main(){ _
    int T, Cas = ;
    cin >> T;
    while(T --){
        cin >> n;
        for(int i = ; i < n; i++) p[i].read();
        double x = ternary_search(, 1e8);
        printf("Case #%d: %.2f %.2f\n", ++Cas, x, cal(x));
    }
}